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Exercises · 8.11

Q.Charcoal pieces of a tree found at an archaeological site have a carbon-14 content that is only 17.5% that of an equivalent sample of carbon from a living tree. What is the age of the tree?

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Step 1. The charcoal's present carbon-14 activity RR is 17.5% of what an equivalent living sample's activity R0R_0 would be: R/R0=0.175R/R_0=0.175.

Step 2. Using R=R0e−λtR=R_0e^{-\lambda t}, rearranged as t=1λln⁡ ⁣(R0R)=1λln⁡ ⁣(10.175)t=\dfrac1\lambda\ln\!\left(\dfrac{R_0}{R}\right)=\dfrac1\lambda\ln\!\left(\dfrac{1}{0.175}\right).

Step 3. The decay constant from the carbon-14 half-life (5730 years) is λ=ln⁡2T1/2=0.69315730≈1.21×10−4 yr−1\lambda=\dfrac{\ln2}{T_{1/2}}=\dfrac{0.6931}{5730}\approx1.21\times10^{-4}\ \text{yr}^{-1}.

Step 4. ln⁡(1/0.175)=ln⁡(5.714)≈1.7430\ln(1/0.175)=\ln(5.714)\approx1.7430.

Step 5. t=1.74301.21×10−4≈1.44×104 yearst=\frac{1.7430}{1.21\times10^{-4}}\approx1.44\times10^4\ \text{years} …

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